\(ĐKXĐ:x\ne2\)

\(P=\dfrac{1}{x-2}-\dfrac{x^2+8}{x^3-8}-\dfrac{4}{x^2+2x+4}\)

\(=\dfrac{x^2+2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{x^2+8}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{4\left(x-2\right)}{x^2+2x+4}\)

\(=\dfrac{x^2+2x+4-x^2-8-4x+8}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{4-2x}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{-2\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=-\dfrac{2}{x^2+2x+4}\)

(2+x).(2-x) = (4-x²)

* hằng đẵng thức đáng nhớ

ý mk là nhân hai cái với nhau  chứ ai cũng bt là = (4-x^2) rồi.

giống kiểu (2+x).(2+x)=4+2x+2x+x^2

a) \(x^2+10x+26+y^2+2y\)

\(=x^2+10x+25+y^2+2y+1\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

b) Xem lại đề 

c) \(x^2-2xy+2y^2+2y+1\)

\(=x^2-2xy+y^2+y^2+2y+1\)

\(=\left(x-y\right)^2+\left(y+1\right)^2\)

ra 900

   \(\dfrac{3}{5}\)a³b x (10ab³ - \(\dfrac{5}{3}\)b² + \(\dfrac{5}{6}\)ab)

=\(\dfrac{3}{5}\)a³b . 10ab³ - \(\dfrac{3}{5}\)a³b. \(\dfrac{5}{3}\)b² + \(\dfrac{3}{5}\)a³b . \(\dfrac{5}{6}\)ab

= 6a⁴b⁴ - a³b³ + \(\dfrac{1}{2}\) a⁴b²

Mn giúp với

thk mn

Lời giải:

$C=(m+n)^3-m^2-2mn-n^2=(m+n)^3-(m^2+2mn+n^2)$

$=(m+n)^3-(m+n)^2=2^3-2^2=8-4=4$

\(C=\left(m+n\right)^3-\left(m^2+2mn+n^2\right)=\left(m+n\right)^3-\left(m+n\right)^2\)

Thay vào ta được 8 - 4 = 4 

đặt:

\(\left\{{}\begin{matrix}S_{ABC}=S^2\\S_{BED}=S_1^2\\S_{CFD}=S_2^2\end{matrix}\right.\)

ta có:

\(ED\) // \(AC\) \(\Rightarrow\Delta BED\sim\Delta BAC\)

\(\Rightarrow\dfrac{S_1^2}{S^2}=\dfrac{BD^2}{BC^2}\Leftrightarrow\dfrac{S_1}{S}=\dfrac{BD}{BC}\)

C/M tương tự:

\(\Rightarrow\dfrac{S_2^2}{S^2}=\dfrac{CD^2}{BC^2}\Leftrightarrow\dfrac{S_2}{S}=\dfrac{CD}{BC}\)

\(\Leftrightarrow\dfrac{BD}{BC}+\dfrac{DC}{BC}=\dfrac{S_1+S_2}{S}\)

\(\Leftrightarrow\dfrac{S_1+S_2}{S}=1\)

\(\Leftrightarrow\left(S_1+S_2\right)^2=S^2\)

\(\Leftrightarrow S^2=\left(\sqrt{100}+\sqrt{16}\right)^2=16^2=256\left(cm^2\right)\)

\(\Leftrightarrow S_{ABC}=256cm^2\)